January 27th, 2012, 03:28 PM  #1 
Member Joined: Jan 2012 Posts: 63 Thanks: 0  continuous
Suppose f:R>R and g:R>R are continuous functions, and define a new function h:R>R by h(x)=max{f(x),g(x)}. Is h continuous? Why? I don't know how to begin; I think I should discuss different situation because f can be greater then g or smaller than g or they have intersection points. 
January 27th, 2012, 10:34 PM  #2 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: continuous
Well, consider the graphs of f and g on the same page. If one function is above the other, say f above g, then h = f and is continuous because f is continuous. If for some x values g is above, for other x values f is above, they must intersect. The intersection will be the max and belong to h. The other parts of h will be from either f or g depending which is on top or other intersections. Ether way h is continuous, though may be composed of different parts of f and g. The point here is that if 2 continuous functions cross each other they must intersect in at least 1 point. Another good question to ask about given the stated conditions...is h(x) differentiable everywhere? Why? 
January 31st, 2012, 02:19 AM  #3  
Senior Member Joined: Apr 2010 Posts: 451 Thanks: 1  Re: continuous Quote:
Take : f(x) = 0 ,which is a continuous function ,and g(x) = sinx , or g(x) = cos x,then the max{f(x),g(x)} are just point above the x axis. Can this function be continuous ??  
January 31st, 2012, 04:27 AM  #4 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: continuous
They are both continuous everywhere but not differentiable everywhere. You lose the derivative whenever they intersect. if f(x)=0 then parts of the x axis belong to h(x), precisely those parts where g(x) <= 0 http://www.wolframalpha.com/input/?i=ma ... ual=Submit http://www.wolframalpha.com/input/?i=ma ... ual=Submit 
January 31st, 2012, 05:17 AM  #5  
Math Team Joined: Apr 2010 Posts: 2,780 Thanks: 361  Re: continuous Quote:
Quote:
Now, h(0) = max{0,sin(0)} is above the xaxis?  

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